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Complex Numbers: Max Distance Between Two Circles = 22 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let the circle C1:∣z∣=rC_1:|z|=r and C2:∣z−3−4i∣=5, z∈CC_2:|z-3-4i|=5,\ z\in\mathbb{C}, be such that C2C_2 lies within C1C_1. If z1z_1 moves on C1C_1, z2z_2 moves on C2C_2 and min⁡∣z1−z2∣=2\min|z_1-z_2|=2, then max⁡∣z1−z2∣\max|z_1-z_2| is equal to
A1212
B1717
C2222correct
D2424
Solution
Step 1: C1C_1 has centre OO and radius ∣z∣=r|z|=r, so OA=rOA=r for a point AA on C1C_1. Step 2: The centre of C2C_2 is 3+4i3+4i, at distance ∣3+4i∣=32+42=5|3+4i|=\sqrt{3^2+4^2}=5 from OO, and its radius is 55. So the farthest point of C2C_2 from OO is at distance 5+5=105+5=10. Step 3: Minimum distance between the circles:
min⁡∣z1−z2∣=r−10=2 ⇒ r=12.\min|z_1-z_2|=r-10=2\ \Rightarrow\ r=12.
 Step 4: Maximum distance is from a point on C1C_1 to the farthest point of C2C_2:
max⁡∣z1−z2∣=r+10=12+10=22.\max|z_1-z_2|=r+10=12+10=22.
 Correct answer: (3)
Solution working
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